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```Discrete Probability Distributions
6- 1
Chapter
Six
McGraw-Hill/Irwin
Random variable
A result from an experiment that, by chance, can
take different values.
Egs.
1. No. of heads you would get in 3 tosses of a coin
2. No. of employees absent on a shift
3. No. of students at CSUN in a semester
4. No. of minutes to drive home from CSUN
5. Inches of rainfall in LA during a year
6. Tire pressure in PSI of a car tire
Examples 1-3 are ‘discrete’
4-6 are ‘continuous’ random variables
In this chapter, we focus on the ‘discrete’.
Probability
Distribution
A listing of all possible outcomes of an
experiment and the corresponding probability.
c
P
F
T
Possible Outcomes
N
= 1/8
= 3/8
= 3/8
= 1/8
Probability Distribution
Note the similarity to histogram
Mean of Discrete Probability Distribution
  [ xP( x)]
where
  represents the mean
o x is each outcome
o P(x) is the probability of the various outcomes x.
Calculating Mean/Expected Value of Heads in 3
coin-toss experiment
  [ xP( x )]
μ = 0 * .125 + 1 * .375 + 2 * .375 + 3 * .125
= 1.5
(which by intuition makes sense because for each toss you have 50%
chance of getting a head).
 is a weighted average.
It is also referred to as Expected Value, E(X).
Binomial Probability Distribution
•An outcome of an experiment is classified into one of
two mutually exclusive categories, such as a success or
failure (bi means two).
•The data collected are the results of counts (hence, a
discrete probability distribution).
•The probability of success stays the same for each trial
(independence).
Binomial Probability Distribution
P( x)n Cx (1   )
x
n x Can you logically
explain this formula?
n is the number of trials
x is the number of observed successes
π is the probability of success on each trial
Cx 
n
n!
x!(n-x)!
Let us re-visit the 3-toss coin experiment
n=3
π = .5
X = 0,1,2,3 (number of heads)
P(0) = 3C0 * (.5)0 * (1 - .5)3-0
P(1) = 3C1 * (.5)1 * (1 - .5)3-1
P(2) = 3C2 * (.5)2 * (1 - .5)3-2
P(3) = 3C3 * (.5)3 * (1 - .5)3-3
= .125
= .375
= .375
= .125
You can also
use
Appendix A
(page 489)
Practice time!
Do Self-Review 6-3, Page 168
Use Appendix A, Page 490
Cumulative Binomial Probability Distribution
Problem 21, Page 173
Use table in Page 491
a. 0.387 ( n=9; x=9; straight from table )
b. 0.001 ( P(x<5) = P(x ≤ 4) )
c. 0.992 ( 1 – P(x ≤ 5) = 1 – 0.008 )
d. 0.946 [ (P(x=7) + (P(x=8) + (P(x=9) ]
Mean of the Binomial Distribution
  n
Logic: If you throw a coin say 100 times, how many
times you would expect to get a head?
For each throw, the probability of getting a head is .5.
So, in 100 trials, you would expect 50 heads (which is
100*.5; ie. nπ )
Variance of the Binomial Distribution
  n (1   )
2
For the previous example,
= 100*.5*(1-.5)
σ 2 = 25
S.D. σ = 5
```
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